Answer
$\begin{aligned}(a) \sigma_{a}=-116.2 \mathrm{MPa} \end{aligned}$
---
$\begin{aligned} (b) \delta_{a}=0.363 \mathrm{mm} \end{aligned}$
Work Step by Step
$$
\Delta T=140-20=120^{\circ} \mathrm{C}
$$
Free thermal expansion:
$$
\begin{aligned} \delta_{T} &=L_{a} \alpha_{a}(\Delta T)+L_{s} \alpha_{s}(\Delta T) \\ &=(0.300)\left(23 \times 10^{-6}\right)(120)+(0.250)\left(17.3 \times 10^{-6}\right)(120) \\ &=1.347 \times 10^{-3} \mathrm{m} \end{aligned}
$$
Shortening due to $P$ to meet constraint:
$\begin{aligned} \delta_{P} &=1.347 \times 10^{-3}-0.5 \times 10^{-3}=0.847 \times 10^{-3} \mathrm{m} \\ \delta_{P} &=\frac{P L_{a}}{E_{a} A_{a}}+\frac{P L_{s}}{E_{s} A_{s}}=\left(\frac{L_{a}}{E_{a} A_{a}}+\frac{L_{s}}{E_{s} A_{s}}\right) P \\ &=\left(\frac{0.300}{\left(75 \times 10^{9}\right)\left(2000 \times 10^{-6}\right)}+\frac{0.250}{\left(190 \times 10^{9}\right)\left(800 \times 10^{-6}\right)}\right) P \\ &=3.6447 \times 10^{-9} P \\ \text { Equating, } & 3.6447 \times 10^{-9} P=0.847 \times 10^{-3} \\ P &=232.39 \times 10^{3} \mathrm{N} \end{aligned}$
$\begin{aligned}(a) & \sigma_{a}=-\frac{P}{A_{a}}=-\frac{232.39 \times 10^{3}}{2000 \times 10^{-6}}=-116.2 \times 10^{6} \mathrm{Pa} \quad \sigma_{a}=-116.2 \mathrm{MPa} \\(b) & \delta_{a}=L_{a} \alpha_{a}(\Delta T)-\frac{P L_{a}}{E_{a} A_{a}} \\ &=(0.300)\left(23 \times 10^{-6}\right)(120)-\frac{\left(232.39 \times 10^{3}\right)(0.300)}{\left(75 \times 10^{9}\right)\left(2000 \times 10^{-6}\right)}=363 \times 10^{-6} \mathrm{m} \quad \delta_{a}=0.363 \mathrm{mm} \end{aligned}$
